# Proving a relation is asymmetrical

I am trying to answer the following:

Consider the relation $T$ on $\mathbb{Z}$ given by $$xTy \Longleftrightarrow x + 1 \le y;$$ Is $xTy$ asymmetric?

$xTy \Rightarrow x\neg Ty$

$xTy \Leftrightarrow x+1 \le y$

$x\neg Ty \Leftrightarrow y+1>y$

We assume that $x \lt y$ is true, then $y+1 \gt x$ will always be true.

Then $xTy$ is asymmetric.

Is this a good enough or even valid proof?

Thanks

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Can you explain what $\,_xNOT_y$ means? – Srivatsan Sep 21 '11 at 17:54
@SrivatsanNarayanan I think he means not related – Ragnar123 Sep 21 '11 at 17:55
You can use $\not T$ or $\lnot T$ if that is what you want. Right click and choose Show Source to see what I wrote. – Ross Millikan Sep 21 '11 at 17:58
Yes, it means Not related, but I did not figure out how to make a not T. - I have updated it now, so it is $¬T$ – Gunnar Sep 21 '11 at 17:58
What kind of objects are $x$ and $y$? This is a relation among what? Integers? Real numbers? $x+1\leq y$ is equivalent to $x\lt y$ for integers, but not for rational or real numbers. $x+1\not\leq y$ is equivalent to $y+1\leq x$ for integers, but not for rational or real numbers. You need to specify not just the "rule" for the relation, but what it is a relation on. – Arturo Magidin Sep 21 '11 at 18:21

A relation $T$ is asymmetrical if and only if whenever $xTy$ holds, $yTx$ does not hold. That is, if and only if: $$xTy\Longrightarrow y(\neg T) x.$$

So what you write makes no sense in terms of proving this. You write that you want to prove that if $xTy$ holds, then $x(\neg T)y$ is true; but this is not what you need to prove.

Also, your definition is that $xTy$ holds if and only if $x+1\leq y$; then you claim that $x(\neg T)y$ is equivalent to $y+1\leq x$; while that may be (see below) a true deduction, it's better to write what the condition means directly: if $xTy$ is equivalent to $x+1\leq y$, then $x(\neg T)y$ is equivalent to $\neg(x+1\leq y)$, which is equivalent to $x+1\not\leq y$.

Now, you don't specify what kind of numbers you are dealing with. For real numbers, $x+1\not\leq y$ is equivalent to $x+1\gt y$; but this is not the same as $y+1\leq x$ if you are dealing with real numbers, or with rational numbers. It's true for integers, but since you don't specify which kinds of $x$s you are dealing with, it could also be incorrect.

I think there are too many errors and unstated assumptions to make what you write a correct proof.

So, is this relation asymmetrical? For real numbers:

We want to prove that if $xTy$, then $y(\neg T)x$. So, assume that $xTy$. That means that $x+1\leq y$. We need to show that $y(\neg T)x$, meaning we need to show that $y+1\not\leq x$.

Indeed, we know that $x\lt x+1\leq y$; therefore, $x\lt y\lt y+1$. By transitivity, if $y+1\leq x$, then we would have $x\lt x$, which is impossible. Therefore, $y+1\not\leq x$. Therefore, $y(\neg T)x$.

Since the relation is asymmetrical for real numbers, it is also asymmetrical for rationals, integers, or natural numbers.

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Thank you very much for your help :-) – Gunnar Sep 21 '11 at 18:37